date()
## [1] "Thu Nov 19 16:18:54 2020"

Week 1, Introduction

Feelings and expectations

How am I feeling? Currently I’m a bit exhausted with all the technical challenges I had with Git and Rstudio integration with several computers. After installing brand new Linux system things were going better though.

So the course started with

  • problems
  • troubles
  • challenges

…but from this on I’m expecting

  • easiness
  • joy
  • wisdom.

I expect to learn R and Git integration, and already did. Looking forward to hear ideas for some data analysis: I have experience with another software and R packages as well, but with R it’s sometimes difficult to see which packages and techniques are supposed or recommended to be used in specific tasks. Also it is interesting to see how approach in data science and conventional statistics differs.

I heard about this course when Kimmo was advertising it via Statnet mailing list.

R calculations…

1+1
## [1] 2
0/0
## [1] NaN

…and inline code.

Square root of two i.e. \(\sqrt 2\) equals 1.4142136.

My GithHub repository:

https://github.com/irual/IODS-project

Week 2, Linear Regression

This week’s exercise is about regression analysis. Tasks include:

Source data: http://s3.amazonaws.com/assets.datacamp.com/production/course_2218/datasets/learning2014.txt.

Starting routines and getting to know data

#Setting working path
setwd("/home/ls/R/projekteja/IODS-project/")

#Reading saved datafile as "excercise2"
exercise2_data <- read.table("./data/ex2.rData")

#Structure of data set (166 obs, 7 vars)
str(exercise2_data)
## 'data.frame':    166 obs. of  7 variables:
##  $ gender  : Factor w/ 2 levels "F","M": 1 2 1 2 2 1 2 1 2 1 ...
##  $ Age     : int  53 55 49 53 49 38 50 37 37 42 ...
##  $ Attitude: int  37 31 25 35 37 38 35 29 38 21 ...
##  $ deep    : num  3.58 2.92 3.5 3.5 3.67 ...
##  $ stra    : num  3.38 2.75 3.62 3.12 3.62 ...
##  $ surf    : num  2.58 3.17 2.25 2.25 2.83 ...
##  $ Points  : int  25 12 24 10 22 21 21 31 24 26 ...
#First 10 observations (showing single string variable, 6 continuous variables), 
head(exercise2_data,n=10)
##    gender Age Attitude     deep  stra     surf Points
## 1       F  53       37 3.583333 3.375 2.583333     25
## 2       M  55       31 2.916667 2.750 3.166667     12
## 3       F  49       25 3.500000 3.625 2.250000     24
## 4       M  53       35 3.500000 3.125 2.250000     10
## 5       M  49       37 3.666667 3.625 2.833333     22
## 6       F  38       38 4.750000 3.625 2.416667     21
## 7       M  50       35 3.833333 2.250 1.916667     21
## 8       F  37       29 3.250000 4.000 2.833333     31
## 9       M  37       38 4.333333 4.250 2.166667     24
## 10      F  42       21 4.000000 3.500 3.000000     26
summary(exercise2_data)
##  gender       Age           Attitude          deep            stra      
##  F:110   Min.   :17.00   Min.   :14.00   Min.   :1.583   Min.   :1.250  
##  M: 56   1st Qu.:21.00   1st Qu.:26.00   1st Qu.:3.333   1st Qu.:2.625  
##          Median :22.00   Median :32.00   Median :3.667   Median :3.188  
##          Mean   :25.51   Mean   :31.43   Mean   :3.680   Mean   :3.121  
##          3rd Qu.:27.00   3rd Qu.:37.00   3rd Qu.:4.083   3rd Qu.:3.625  
##          Max.   :55.00   Max.   :50.00   Max.   :4.917   Max.   :5.000  
##       surf           Points     
##  Min.   :1.583   Min.   : 7.00  
##  1st Qu.:2.417   1st Qu.:19.00  
##  Median :2.833   Median :23.00  
##  Mean   :2.787   Mean   :22.72  
##  3rd Qu.:3.167   3rd Qu.:27.75  
##  Max.   :4.333   Max.   :33.00
#Gender proportions: females 66.3%, males 33.7%
prop.table(table(exercise2_data$gender))
## 
##         F         M 
## 0.6626506 0.3373494

So that was basic information about data set, which consist of 166 observations and single factor, six continuous variables, with 66% of subjects being female. For more information, please visit http://www.helsinki.fi/~kvehkala/JYTmooc/JYTOPKYS3-meta.txt where the original data set described.

Visualizing pairwise distributions

#Loading additional packages and plotting bivariate distributions 
library(ggplot2)
library(GGally) 

ggpairs(exercise2_data, aes(col=gender, alpha=0.3), 
        upper=list(continuous = wrap("cor", size=2.5)),
        lower=list(combo=wrap("facethist", bins=25))) +
  scale_fill_manual(values = c("red","blue"))

In graph above, red color indicates female gender, blue male. Correlation text size decreased for better fit.

Highest absolute correlation coefficient with Points are:

Variables Pearson R
Attitude * Points 0.437
stra * Points 0.146
surf * Points -0.144

-> Let’s select these three as predictors. Points is the outcome.

Multiple linear regression model #1

glm1 <- lm(Points ~ Attitude + stra + surf, data=exercise2_data)
summary(glm1)
## 
## Call:
## lm(formula = Points ~ Attitude + stra + surf, data = exercise2_data)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -17.1550  -3.4346   0.5156   3.6401  10.8952 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 11.01711    3.68375   2.991  0.00322 ** 
## Attitude     0.33952    0.05741   5.913 1.93e-08 ***
## stra         0.85313    0.54159   1.575  0.11716    
## surf        -0.58607    0.80138  -0.731  0.46563    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 5.296 on 162 degrees of freedom
## Multiple R-squared:  0.2074, Adjusted R-squared:  0.1927 
## F-statistic: 14.13 on 3 and 162 DF,  p-value: 3.156e-08

Attitude is the only statistically significant predictor for Points (p<0.001).

Surf is least significant predictor: let’s remove it from the model.

Multiple linear regression model #2

glm2 <- lm(Points ~ Attitude + stra, data=exercise2_data)
summary(glm2)
## 
## Call:
## lm(formula = Points ~ Attitude + stra, data = exercise2_data)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -17.6436  -3.3113   0.5575   3.7928  10.9295 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  8.97290    2.39591   3.745  0.00025 ***
## Attitude     0.34658    0.05652   6.132 6.31e-09 ***
## stra         0.91365    0.53447   1.709  0.08927 .  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 5.289 on 163 degrees of freedom
## Multiple R-squared:  0.2048, Adjusted R-squared:  0.1951 
## F-statistic: 20.99 on 2 and 163 DF,  p-value: 7.734e-09

Attitude is still the only statistically significant predictor for Points (p<0.001).

Stra is not significant, but still p<0.1. We might leave the model as it is, but out of curiosity, let’s remove Stra anyway and see what happens.

Simple linear regression model

glm3 <- lm(Points ~ Attitude, data=exercise2_data)
summary(glm3)
## 
## Call:
## lm(formula = Points ~ Attitude, data = exercise2_data)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -16.9763  -3.2119   0.4339   4.1534  10.6645 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 11.63715    1.83035   6.358 1.95e-09 ***
## Attitude     0.35255    0.05674   6.214 4.12e-09 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 5.32 on 164 degrees of freedom
## Multiple R-squared:  0.1906, Adjusted R-squared:  0.1856 
## F-statistic: 38.61 on 1 and 164 DF,  p-value: 4.119e-09

Model parameter estimates interpretation:

Attitude is still highly significant predictor for Points (p<0.001).

Intercept equals 11.6. This is the value of Points where Attitude=0, based on model formula.

Slope parameter of Attitude equals to 0.352, i.e. single unit increase in Attitude reflects as 0.352 increase to Points.

Coefficents of determination: unadjusted 19.1%, adjusted 18.6%. Both are a bit lower than in previous model. But let’s leave it that way.

Diagnostic plots for model #3

#Drawing diagnostic plots. Choosing the plots 1, 2 and 5. Full list of plot options:
#1. Residual vs fitted
#2. Normal QQ plot  
#3. Scale-location  
#4. Cooks distance
#5. Residuals vs. Leverage
#6. Cooks distances vs Leverage

#Defining plot matrix of 1x3. Pin defines 1:1 aspect ratio for each sub plot.
par(mfrow = c(1,3), pin=c(1.75,1.75))
plot(glm3, which=c(1,2,5))

Residual vs. fitted plot: Distibution is symmetric, there are no outliers. Looks fine.

Normal Quantile-Quantile plot: There is slight deviation from the diagonal at the ends. However, this is still not yet alarming at all.

Residual vs. Leverage plot: Highest Cook’s distance value is <0.05, i.e. low. No problems here.

Conclusion: model seems to be adequate, assumptions are met, based on visual examination. Attitude predicts value of Points.

Week 3, Logistic Regression

This week’s exercise is about logistic regression analysis. Tasks include:

Source data: Secondary school student alcohol consumption in Portugal (P. Cortez and A. Silva. Using Data Mining to Predict Secondary School Student Performance). Center for Machine Learning and Intelligent Systems at the University of California, Irvine.

Data: https://archive.ics.uci.edu/ml/datasets/Student+Performance.

Description: https://archive.ics.uci.edu/ml/datasets/Student+Performance#.

2. Starting routines and getting to know data

#Setting working path
setwd("/home/ls/R/projekteja/IODS-project/")

#Reading the data created with "creatle_alc.R".
alc <- read.table("./data/alc.rData")

Loading library dplyr for further DM needs.

library(dplyr)
## 
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
## 
##     filter, lag
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union
#Variable dimensions, variable names and some values.
glimpse(alc)
## Rows: 370
## Columns: 51
## $ school     <fct> GP, GP, GP, GP, GP, GP, GP, GP, GP, GP, GP, GP, GP, GP, GP…
## $ sex        <fct> F, F, F, F, F, F, F, F, F, F, F, F, F, F, F, F, F, F, F, F…
## $ age        <int> 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15…
## $ address    <fct> R, R, R, R, R, R, R, R, R, U, U, U, U, U, U, U, U, U, U, U…
## $ famsize    <fct> GT3, GT3, GT3, GT3, GT3, GT3, GT3, LE3, LE3, GT3, GT3, GT3…
## $ Pstatus    <fct> T, T, T, T, T, T, T, T, T, A, A, T, T, T, T, T, T, T, T, T…
## $ Medu       <int> 1, 1, 2, 2, 3, 3, 3, 2, 3, 3, 4, 1, 1, 1, 1, 1, 2, 2, 2, 3…
## $ Fedu       <int> 1, 1, 2, 4, 3, 4, 4, 2, 1, 3, 3, 1, 1, 1, 2, 2, 1, 2, 3, 2…
## $ Mjob       <fct> at_home, other, at_home, services, services, services, ser…
## $ Fjob       <fct> other, other, other, health, services, health, teacher, se…
## $ reason     <fct> home, reputation, reputation, course, reputation, course, …
## $ guardian   <fct> mother, mother, mother, mother, other, mother, father, mot…
## $ traveltime <int> 2, 1, 1, 1, 2, 1, 2, 2, 2, 1, 1, 3, 1, 1, 1, 1, 3, 1, 2, 1…
## $ studytime  <int> 4, 2, 1, 3, 3, 3, 3, 2, 4, 4, 2, 1, 2, 2, 2, 2, 3, 4, 1, 2…
## $ schoolsup  <fct> yes, yes, yes, yes, no, yes, no, yes, no, yes, no, no, no,…
## $ famsup     <fct> yes, yes, yes, yes, yes, yes, yes, yes, yes, no, yes, yes,…
## $ activities <fct> yes, no, yes, yes, yes, yes, no, no, no, no, yes, yes, yes…
## $ nursery    <fct> yes, no, yes, yes, yes, yes, yes, yes, no, yes, yes, no, n…
## $ higher     <fct> yes, yes, yes, yes, yes, yes, yes, yes, yes, yes, yes, yes…
## $ internet   <fct> yes, yes, no, yes, yes, yes, yes, yes, yes, no, yes, yes, …
## $ romantic   <fct> no, yes, no, no, yes, no, yes, no, no, no, no, yes, no, no…
## $ famrel     <int> 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 3, 5, 5, 3, 3…
## $ freetime   <int> 1, 3, 3, 3, 2, 3, 2, 1, 4, 3, 3, 3, 3, 4, 3, 2, 2, 1, 5, 3…
## $ goout      <int> 2, 4, 1, 2, 1, 2, 2, 3, 2, 3, 2, 3, 2, 2, 2, 3, 2, 2, 1, 2…
## $ Dalc       <int> 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1…
## $ Walc       <int> 1, 4, 1, 1, 3, 1, 2, 3, 3, 1, 1, 2, 3, 2, 1, 2, 1, 1, 1, 1…
## $ health     <int> 1, 5, 2, 5, 3, 5, 5, 4, 3, 4, 1, 4, 4, 5, 5, 1, 4, 3, 5, 3…
## $ n          <int> 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2…
## $ id.p       <int> 1096, 1073, 1040, 1025, 1166, 1039, 1131, 1069, 1070, 1106…
## $ id.m       <int> 2096, 2073, 2040, 2025, 2153, 2039, 2131, 2069, 2070, 2106…
## $ failures   <int> 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2…
## $ paid       <fct> yes, no, no, no, yes, no, no, no, no, no, no, no, no, no, …
## $ absences   <int> 3, 2, 8, 2, 5, 2, 0, 1, 9, 10, 0, 3, 2, 0, 4, 1, 2, 6, 2, …
## $ G1         <int> 10, 10, 14, 10, 12, 12, 11, 10, 16, 10, 14, 10, 11, 10, 12…
## $ G2         <int> 12, 8, 13, 10, 12, 12, 6, 10, 16, 10, 14, 6, 11, 12, 12, 1…
## $ G3         <int> 12, 8, 12, 9, 12, 12, 6, 10, 16, 10, 15, 6, 11, 12, 12, 14…
## $ failures.p <int> 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1…
## $ paid.p     <fct> yes, no, no, no, yes, no, no, no, no, no, no, no, no, no, …
## $ absences.p <int> 4, 2, 8, 2, 2, 2, 0, 0, 6, 10, 0, 6, 2, 0, 6, 0, 0, 4, 4, …
## $ G1.p       <int> 13, 13, 14, 10, 13, 11, 10, 11, 15, 10, 15, 11, 13, 12, 13…
## $ G2.p       <int> 13, 11, 13, 11, 13, 12, 11, 10, 15, 10, 14, 12, 12, 12, 12…
## $ G3.p       <int> 13, 11, 12, 10, 13, 12, 12, 11, 15, 10, 15, 13, 12, 12, 13…
## $ failures.m <int> 1, 2, 0, 0, 2, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3…
## $ paid.m     <fct> yes, no, yes, yes, yes, yes, no, yes, no, no, yes, no, no,…
## $ absences.m <int> 2, 2, 8, 2, 8, 2, 0, 2, 12, 10, 0, 0, 2, 0, 2, 2, 4, 8, 0,…
## $ G1.m       <int> 7, 8, 14, 10, 10, 12, 12, 8, 16, 10, 14, 8, 9, 8, 10, 16, …
## $ G2.m       <int> 10, 6, 13, 9, 10, 12, 0, 9, 16, 11, 15, 0, 10, 11, 11, 15,…
## $ G3.m       <int> 10, 5, 13, 8, 10, 11, 0, 8, 16, 11, 15, 0, 10, 11, 11, 15,…
## $ alc_use    <dbl> 1.0, 3.0, 1.0, 1.0, 2.5, 1.0, 2.0, 2.0, 2.5, 1.0, 1.0, 1.5…
## $ high_use   <lgl> FALSE, TRUE, FALSE, FALSE, TRUE, FALSE, FALSE, FALSE, TRUE…
## $ cid        <int> 3001, 3002, 3003, 3004, 3005, 3006, 3007, 3008, 3009, 3010…

So there are 370 cases and 51 variables. We’ll concentrate on
- alc_use, which is derived as a mean of Dalc and Walc (workday and weekend alcohol consumption, varying from 1 - very low to 5 - very high)
- high_use, which is dichotomization of alc_use with cut-point 2.

3. Selecting variables for further analysis

Choosing four interesting variables for further analysis. Response variable will be high/low alcohol consumpition.

Interesting/potential predictors I chose:
1. sex - student’s sex (binary: ‘F’ - female or ‘M’ - male). Gender will definitely has an effect on alcohol consumption. On average males drink more. We’ll see how it goes here.
2. internet - Internet access at home (binary: yes or no). Data is from 2014. Could be possible that subjects with internet connection are skulking at their homes instead of having a good time in restaurants etc. So internet access may decrease alcohol consumption. Maybe.
3. romantic - with a romantic relationship (binary: yes or no). Having a spouse may cause a lot of stress which leads to drinking (helping or not). On the other hand, relationship may decrease drinking (all time goes sitting hand in hand or spouse may prohibit alcohol use. Who knows, there are several possible associations.
4. absences - number of school absences (numeric: from 0 to 93). Drinking may cause school absences. Or there can be some common factor causing both absences and drinking. Caution: absences is count variable while other ones are dichotomous.

These predictors may have an effect, but after all it’s hard to know beforehand. Let’s explore variables and associations.

4. Exploring associations of alchol use and predictor. Tables and plots.

#Loading ggplot2 for better plotting opportunities
library(ggplot2)

g1 <- ggplot(data = alc, aes(x = high_use))

g1 + geom_bar(aes(fill=sex))

g1 + geom_bar(aes(fill=internet))

g1 + geom_bar(aes(fill=romantic))

g1 + geom_boxplot(aes(y=absences, fill=high_use)) +
  stat_summary(fun=mean, geom="point", aes(y=absences), col="blue", size=5, shape="diamond") +
  theme(legend.position="none")

Those were the graphs. Gender distribution seems to be a bit different in alcohol groups. Mean absence counts seems to higher in high use group (mean=blue diamond symbol). Absence distributions seems to be higly negatively skew, but at least distributions are similary shaped in both group. This is not perfect situation, but usually logistic regression tolerates quite well such phenomenon.

One-way table and crosstabulation of alcohol use vs. predictors. Original counts.

table("High alcohol use"=alc$high_use)
## High alcohol use
## FALSE  TRUE 
##   259   111
table("High alcohol use"=alc$high_use, "Sex"=alc$sex)
##                 Sex
## High alcohol use   F   M
##            FALSE 154 105
##            TRUE   41  70
table("High alcohol use"=alc$high_use, "Internet access at home (binary: yes or no)"=alc$internet)
##                 Internet access at home (binary: yes or no)
## High alcohol use  no yes
##            FALSE  42 217
##            TRUE   15  96
table("High alcohol use"=alc$high_use, "Romantic relationship (binary: yes or no)"=alc$romantic)
##                 Romantic relationship (binary: yes or no)
## High alcohol use  no yes
##            FALSE 173  86
##            TRUE   78  33
#Basic descriptives by group
tapply(alc$absences, alc$high_use, summary)
## $`FALSE`
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##    0.00    1.00    3.00    3.71    5.00   45.00 
## 
## $`TRUE`
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   0.000   2.000   4.000   6.378   9.000  44.000

Response counts are 259 (low) and 111 (high alcohol consumption). Crosstabs shows the same numbers as the previous plots. Abscense has higher mean in high alcohol use group (6.4 vs. 3.7).

Let’s have percentages (or proportions, to be exact) where figures are scaled for each row sum to be 100%. This can be achieved using pipe and function call prop.table(margin=1).

table("High alcohol use"=alc$high_use, "Sex"=alc$sex) %>% prop.table(margin=1)
##                 Sex
## High alcohol use         F         M
##            FALSE 0.5945946 0.4054054
##            TRUE  0.3693694 0.6306306
table("High alcohol use"=alc$high_use, "Internet access at home (binary: yes or no)"=alc$internet) %>% prop.table(margin=1)
##                 Internet access at home (binary: yes or no)
## High alcohol use        no       yes
##            FALSE 0.1621622 0.8378378
##            TRUE  0.1351351 0.8648649
table("High alcohol use"=alc$high_use, "Romantic relationship (binary: yes or no)"=alc$romantic) %>% prop.table(margin=1)
##                 Romantic relationship (binary: yes or no)
## High alcohol use        no       yes
##            FALSE 0.6679537 0.3320463
##            TRUE  0.7027027 0.2972973

Proportion of males is higher in high alcohol consumption group. Gender proportions in internet access and romantic relationship groups are pretty much the same.

5. Logistic regression analysis

General logistic model with multiple predictors can be defined as \[ \log\left(\frac{p({\bf x})}{1 - p({\bf x})}\right) = \beta_0 + \beta_1 x_1 + \ldots + \beta_{p - 1} x_{p - 1} \]

Let’s fit binary response logistic model with multiple predictors with glm function. Outcome is high_use with TRUE indicating high alcohol consumption as an event.

# find the model with glm()
# Model 1 (full model)
m1 <- glm(high_use ~ sex + internet + romantic + absences, data = alc, family = "binomial")
summary(m1)
## 
## Call:
## glm(formula = high_use ~ sex + internet + romantic + absences, 
##     family = "binomial", data = alc)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.2366  -0.8749  -0.6095   1.1165   2.0636  
## 
## Coefficients:
##              Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.791835   0.354025  -5.061 4.16e-07 ***
## sexM         1.022570   0.245172   4.171 3.04e-05 ***
## internetyes  0.006198   0.340680   0.018    0.985    
## romanticyes -0.217035   0.263397  -0.824    0.410    
## absences     0.098186   0.023536   4.172 3.02e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 452.04  on 369  degrees of freedom
## Residual deviance: 414.97  on 365  degrees of freedom
## AIC: 424.97
## 
## Number of Fisher Scoring iterations: 4

Access to internet doesn’t seem to be statistically significant predictor for high alcohol use. Let’s remove it from the model and try again.

#Model 2
m2 <- glm(high_use ~ sex + romantic + absences, data = alc, family = "binomial")
summary(m2)
## 
## Call:
## glm(formula = high_use ~ sex + romantic + absences, family = "binomial", 
##     data = alc)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.2370  -0.8745  -0.6091   1.1168   2.0641  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.78707    0.23822  -7.502 6.30e-14 ***
## sexM         1.02291    0.24444   4.185 2.86e-05 ***
## romanticyes -0.21668    0.26268  -0.825    0.409    
## absences     0.09823    0.02341   4.196 2.72e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 452.04  on 369  degrees of freedom
## Residual deviance: 414.97  on 366  degrees of freedom
## AIC: 422.97
## 
## Number of Fisher Scoring iterations: 4

Couple relationship either doesn’t seem to be statistically significant predictor for high alcohol. Let’s remove it from the model as well and try again.

#Model 3 (final model)
m3 <- glm(high_use ~ sex + absences, data = alc, family = "binomial")
summary(m3)
## 
## Call:
## glm(formula = high_use ~ sex + absences, family = "binomial", 
##     data = alc)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.2706  -0.8838  -0.5901   1.0960   1.9993  
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.85303    0.22609  -8.196 2.49e-16 ***
## sexM         1.03451    0.24395   4.241 2.23e-05 ***
## absences     0.09671    0.02336   4.140 3.48e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 452.04  on 369  degrees of freedom
## Residual deviance: 415.66  on 367  degrees of freedom
## AIC: 421.66
## 
## Number of Fisher Scoring iterations: 4

Now both remaining factors are statistically significant (p<0.001), we can leave the model as it is. Slope parameter for Sex (being male as an event) is 1.03, with standard error 0.24, for the event of alcohol consumption being high. However, this is log-odds, which may be difficult to interpret. Therefor, let’s calculate and show Odds Ratios which are exponentiated versions of shown estimates, i.e. constant e is raised to power of that number.

This is somewhat plausible result. I’m still surprised that internet access and couple relationsip didn’t have any significant role.

# compute odds ratios (OR)
OR <- coef(m3) %>% exp

# compute confidence intervals (CI)
CI <- confint(m3) %>% exp
## Waiting for profiling to be done...
# print out the odds ratios with their confidence intervals
cbind(OR, CI)
##                    OR      2.5 %    97.5 %
## (Intercept) 0.1567616 0.09877521 0.2401977
## sexM        2.8137400 1.75518106 4.5760120
## absences    1.1015380 1.05453436 1.1557571

So the Odds Ratio (95% Confidence Interval) for Sex is 2.81 (1.76-4.58) and for absences 1.10 (1.05-1.16). Males has 2.8-fold odds of becoming to high alcohol consumption group compared to females. Similarly, single unit increase in absence count scale makes odds of being in high alcohol consumption group grow 1.1-fold, i.e. it increases 10%.

Just to be sure, let’s fit single predictor model and compare manual OR calculation with R model results.

#Model 4 (single predictor model)
m4 <- glm(high_use ~ sex, data = alc, family = "binomial")
OR <- coef(m4) %>% exp
CI <- confint(m4) %>% exp
## Waiting for profiling to be done...
cbind(OR, CI)
##                    OR     2.5 %    97.5 %
## (Intercept) 0.2662338 0.1863238 0.3717646
## sexM        2.5040650 1.5902950 3.9838892
# tabulate the target variable versus the predictor
table(high_use = alc$high_use, alc$sex) 
##         
## high_use   F   M
##    FALSE 154 105
##    TRUE   41  70

OR for sex in single predictor model was 2.5040650. We are supposed to get the same result when calculating males/girls ratio in high vs. low alcohol consumption group, i.e.

\[ \frac{154/105} {41/70} \]

#Calculate with R
(154/105)/(41/70)  
## [1] 2.504065

Yes, it works.

6. Studying predictive power of logistic regression model

Let’s see how previous model #3 is able predict alcohol usage.

Providing a 2x2 cross tabulation of predictions versus the actual values. First we need to predict values, add them into data set, categorize results and peep the data.

# predict() the probability of high_use
probabilities <- predict(m3, type = "response")

# add the predicted probabilities to 'alc'
alc <- mutate(alc, probability = probabilities)

# use the probabilities to make a prediction of high_use
alc <- mutate(alc, prediction = probability > 0.5)

# see the last ten original classes, predicted probabilities, and class predictions
select(alc, sex, probability, high_use, prediction) %>% tail(15)
##     sex probability high_use prediction
## 356   M   0.3708919     TRUE      FALSE
## 357   M   0.3937240     TRUE      FALSE
## 358   M   0.3486225     TRUE      FALSE
## 359   M   0.3486225    FALSE      FALSE
## 360   M   0.5129601     TRUE       TRUE
## 361   M   0.4646683     TRUE      FALSE
## 362   M   0.3708919     TRUE      FALSE
## 363   M   0.3486225     TRUE      FALSE
## 364   M   0.3937240     TRUE      FALSE
## 365   M   0.3708919    FALSE      FALSE
## 366   M   0.3937240     TRUE      FALSE
## 367   M   0.3060791    FALSE      FALSE
## 368   M   0.3937240     TRUE      FALSE
## 369   M   0.4887880     TRUE      FALSE
## 370   M   0.3060791    FALSE      FALSE

Now we are ready to see how predictions matches the reality.

# tabulate the target variable versus the predictions
table(high_use = alc$high_use, prediction = alc$prediction) 
##         prediction
## high_use FALSE TRUE
##    FALSE   251    8
##    TRUE     86   25

So, in high alcohol use group 86 cases are correctly predicted to be in that group, 25 cases are predicted to wrong group. In low alc. group prediction is correct in 251 case and incorrect in 8 case. Let’s see percentages.

table(high_use = alc$high_use, prediction = alc$prediction) %>% prop.table(margin=1)
##         prediction
## high_use      FALSE       TRUE
##    FALSE 0.96911197 0.03088803
##    TRUE  0.77477477 0.22522523

In low alc. group prediction goes right in 97% cases, in high group prediction is correct only in 23% of cases. Clearly model is prone to find cases of low alcohol consumption, but has difficulties to find users of higher level of alcohol.

Let’s plot the results.

# Some data management. Proportions into data frame.
props <- as.data.frame(table(high_use = alc$high_use, prediction = alc$prediction) %>% prop.table(margin=1))

# Plotting with ggplot2
ggplot(data = props) +
  geom_bar(stat="identity", aes(x=high_use, y=Freq, fill=prediction)) +
  scale_y_continuous(labels=scales::percent) +
  ylab("Proportion")

# Plotting with ggplot2
ggplot(data = alc) +
  geom_point(aes(x = probability, y = high_use, color = prediction), size=5, alpha=0.5) +
  ylab("High alcohol use")

This shows the same story. With selected probability 50% model will catch most low users, but is not able to detect most cases in high alcohol use group. Maybe prob. threshold of 50% should be raised?

10-fold cross-validation (“bonus task”)

Let’s perform 10-fold cross-validation on our model and see if it has better test set performance (smaller prediction error using 10-fold cross-validation) compared to the model introduced in DataCamp (which had about 0.26 error).

# define a loss function (mean prediction error)
loss_func <- function(class, prob) {
  n_wrong <- abs(class - prob) > 0.5
  mean(n_wrong)
}

# compute the average number of wrong predictions in the (training) data
loss_func(class = alc$high_use, prob = alc$probability)
## [1] 0.2540541
library(boot)
#10-fold cross-validation
cv <- cv.glm(data = alc, cost = loss_func, glmfit = m3, K = 10)

# average number of wrong predictions in the cross validation
cv$delta[1]
## [1] 0.2621622

So this is only slightly lower prediction error than in DataCamp excercise - the same in practise.

10-fold cross-validation comparison (“super-bonus task”)

Let’s conduct comparison of several cross-validations with different sets of predictors. Starting with a very high number of predictors and exploring the changes in the training and testing errors as moving to models with less predictors.

There was some problems with conducting analysis with huge number of analysis, estimation did not converge. Se let’s start with single predictor model and add predictors one by one until we’re in 20.

I first tried to do the whole thing with R functions, letting R to generate all models, validations and error estimates automatically. However, I wasn’t clever enough to create lists of predictors for glm without additional quotes, so I was stuck. Now modelling etc. is done manually model by model, which is akward…

m1  <- glm(high_use ~ school, data = alc, family = "binomial")
m2  <- glm(high_use ~ school + sex + age, data = alc, family = "binomial")
m3  <- glm(high_use ~ school + sex + age + address, data = alc, family = "binomial")
m4  <- glm(high_use ~ school + sex + age + address + famsize, data = alc, family = "binomial")
m5  <- glm(high_use ~ school + sex + age + address + famsize + Pstatus + Medu, data = alc, family = "binomial")
m6  <- glm(high_use ~ school + sex + age + address + famsize + Pstatus + Medu + Fedu, data = alc, family = "binomial")
m7  <- glm(high_use ~ school + sex + age + address + famsize + Pstatus + Medu + Fedu + Mjob, data = alc, family = "binomial")
m8  <- glm(high_use ~ school + sex + age + address + famsize + Pstatus + Medu + Fedu + Mjob + Fjob, data = alc, family = "binomial")
m9  <- glm(high_use ~ school + sex + age + address + famsize + Pstatus + Medu + Fedu + Mjob + Fjob + reason, data = alc, family = "binomial")
m10 <- glm(high_use ~ school + sex + age + address + famsize + Pstatus + Medu + Fedu + Mjob + Fjob + reason + guardian, data = alc, family = "binomial")
m11 <- glm(high_use ~ school + sex + age + address + famsize + Pstatus + Medu + Fedu + Mjob + Fjob + reason + guardian + traveltime, data = alc, family = "binomial")
m12 <- glm(high_use ~ school + sex + age + address + famsize + Pstatus + Medu + Fedu + Mjob + Fjob + reason + guardian + traveltime + studytime, data = alc, family = "binomial")
m13 <- glm(high_use ~ school + sex + age + address + famsize + Pstatus + Medu + Fedu + Mjob + Fjob + reason + guardian + traveltime + studytime + schoolsup, data = alc, family = "binomial")
m14 <- glm(high_use ~ school + sex + age + address + famsize + Pstatus + Medu + Fedu + Mjob + Fjob + reason + guardian + traveltime + studytime + schoolsup + famsup, data = alc, family = "binomial")
m15 <- glm(high_use ~ school + sex + age + address + famsize + Pstatus + Medu + Fedu + Mjob + Fjob + reason + guardian + traveltime + studytime + schoolsup + famsup + activities, data = alc, family = "binomial")
m16 <- glm(high_use ~ school + sex + age + address + famsize + Pstatus + Medu + Fedu + Mjob + Fjob + reason + guardian + traveltime + studytime + schoolsup + famsup + activities + nursery, data = alc, family = "binomial")
m17 <- glm(high_use ~ school + sex + age + address + famsize + Pstatus + Medu + Fedu + Mjob + Fjob + reason + guardian + traveltime + studytime + schoolsup + famsup + activities + nursery + higher, data = alc, family = "binomial")
m18 <- glm(high_use ~ school + sex + age + address + famsize + Pstatus + Medu + Fedu + Mjob + Fjob + reason + guardian + traveltime + studytime + schoolsup + famsup + activities + nursery + higher + internet, data = alc, family = "binomial")
m19 <- glm(high_use ~ school + sex + age + address + famsize + Pstatus + Medu + Fedu + Mjob + Fjob + reason + guardian + traveltime + studytime + schoolsup + famsup + activities + nursery + higher + internet + romantic, data = alc, family = "binomial")
m20 <- glm(high_use ~ school + sex + age + address + famsize + Pstatus + Medu + Fedu + Mjob + Fjob + reason + guardian + traveltime + studytime + schoolsup + famsup + activities + nursery + higher + internet + romantic + famrel, data = alc, family = "binomial")

That was the modeling. Next, let’s conduct 10-fold cross-validation for each model.

cv1  <- cv.glm(data = alc, cost = loss_func, glmfit = m1,  K = 10)
cv2  <- cv.glm(data = alc, cost = loss_func, glmfit = m2,  K = 10)
cv3  <- cv.glm(data = alc, cost = loss_func, glmfit = m3,  K = 10)
cv4  <- cv.glm(data = alc, cost = loss_func, glmfit = m4,  K = 10)
cv5  <- cv.glm(data = alc, cost = loss_func, glmfit = m5,  K = 10)
cv6  <- cv.glm(data = alc, cost = loss_func, glmfit = m6,  K = 10)
cv7  <- cv.glm(data = alc, cost = loss_func, glmfit = m7,  K = 10)
cv8  <- cv.glm(data = alc, cost = loss_func, glmfit = m8,  K = 10)
cv9  <- cv.glm(data = alc, cost = loss_func, glmfit = m9,  K = 10)
cv10 <- cv.glm(data = alc, cost = loss_func, glmfit = m10, K = 10)
cv11 <- cv.glm(data = alc, cost = loss_func, glmfit = m11, K = 10)
cv12 <- cv.glm(data = alc, cost = loss_func, glmfit = m12, K = 10)
cv13 <- cv.glm(data = alc, cost = loss_func, glmfit = m13, K = 10)
cv14 <- cv.glm(data = alc, cost = loss_func, glmfit = m14, K = 10)
cv15 <- cv.glm(data = alc, cost = loss_func, glmfit = m15, K = 10)
cv16 <- cv.glm(data = alc, cost = loss_func, glmfit = m16, K = 10)
cv17 <- cv.glm(data = alc, cost = loss_func, glmfit = m17, K = 10)
cv18 <- cv.glm(data = alc, cost = loss_func, glmfit = m18, K = 10)
cv19 <- cv.glm(data = alc, cost = loss_func, glmfit = m19, K = 10)
cv20 <- cv.glm(data = alc, cost = loss_func, glmfit = m20, K = 10)

Let’s create a data frame comprising deltas and number of predictors.

deltas <- c(cv1$delta[1],  cv2$delta[1],  cv3$delta[1],  cv4$delta[1],  cv5$delta[1],
            cv6$delta[1],  cv7$delta[1],  cv8$delta[1],  cv9$delta[1],  cv10$delta[1],
            cv11$delta[1], cv12$delta[1], cv13$delta[1], cv14$delta[1], cv15$delta[1],
            cv16$delta[1], cv17$delta[1], cv18$delta[1], cv19$delta[1], cv20$delta[1])

preds <- c(1:20)

compdata <- data.frame(deltas,preds)

And plotting the results.

ggplot(compdata, aes(x=preds, y=deltas)) +
  geom_line()

We’ll see that average amount of wrong predictions in the cross validation is not increasing/decreasing linearly. However, looks like error is mostly increasing when more complicated models are fitted. Simple models are more preferable.

Week 4, Clustering and Classification

This week’s exercise is about cluster analysis and classification. Tasks include:

2. Starting routines and getting to know data

Setting working path, loading library and data set.

#Setting working path
setwd("/home/ls/R/projekteja/IODS-project/")

# access the MASS package
library(MASS)
## 
## Attaching package: 'MASS'
## The following object is masked from 'package:dplyr':
## 
##     select
# load the data
data("Boston")

Boston data is included in the MASS package. Data is about housing values in suburbs of Boston city, with other demographic and environmental information as well. Further information: https://stat.ethz.ch/R-manual/R-devel/library/MASS/html/Boston.html.

List of variables (from above url):

Variable name Definition
crim per capita crime rate by town.
zn proportion of residential land zoned for lots over 25,000 sq.ft.
indus proportion of non-retail business acres per town.
chas Charles River dummy variable (= 1 if tract bounds river; 0 otherwise).
nox nitrogen oxides concentration (parts per 10 million).
rm average number of rooms per dwelling.
age proportion of owner-occupied units built prior to 1940.
dis weighted mean of distances to five Boston employment centres.
rad index of accessibility to radial highways.
tax full-value property-tax rate per $10,000.
ptratio pupil-teacher ratio by town.
black 1000(Bk - 0.63)\(^{2}\) where Bk is the proportion of blacks by town.
lstat lower status of the population (percent).
medv median value of owner-occupied homes in $1000s.

Sources:
- Harrison, D. and Rubinfeld, D.L. (1978) Hedonic prices and the demand for clean air. J. Environ. Economics and Management 5, 81–102.
- Belsley D.A., Kuh, E. and Welsch, R.E. (1980) Regression Diagnostics. Identifying Influential Data and Sources of Collinearity. New York: Wiley.

Loading library dplyr for further DM needs.

library(dplyr) 

#Data dimensions, variable names and some values.
glimpse(Boston)
## Rows: 506
## Columns: 14
## $ crim    <dbl> 0.00632, 0.02731, 0.02729, 0.03237, 0.06905, 0.02985, 0.08829…
## $ zn      <dbl> 18.0, 0.0, 0.0, 0.0, 0.0, 0.0, 12.5, 12.5, 12.5, 12.5, 12.5, …
## $ indus   <dbl> 2.31, 7.07, 7.07, 2.18, 2.18, 2.18, 7.87, 7.87, 7.87, 7.87, 7…
## $ chas    <int> 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0…
## $ nox     <dbl> 0.538, 0.469, 0.469, 0.458, 0.458, 0.458, 0.524, 0.524, 0.524…
## $ rm      <dbl> 6.575, 6.421, 7.185, 6.998, 7.147, 6.430, 6.012, 6.172, 5.631…
## $ age     <dbl> 65.2, 78.9, 61.1, 45.8, 54.2, 58.7, 66.6, 96.1, 100.0, 85.9, …
## $ dis     <dbl> 4.0900, 4.9671, 4.9671, 6.0622, 6.0622, 6.0622, 5.5605, 5.950…
## $ rad     <int> 1, 2, 2, 3, 3, 3, 5, 5, 5, 5, 5, 5, 5, 4, 4, 4, 4, 4, 4, 4, 4…
## $ tax     <dbl> 296, 242, 242, 222, 222, 222, 311, 311, 311, 311, 311, 311, 3…
## $ ptratio <dbl> 15.3, 17.8, 17.8, 18.7, 18.7, 18.7, 15.2, 15.2, 15.2, 15.2, 1…
## $ black   <dbl> 396.90, 396.90, 392.83, 394.63, 396.90, 394.12, 395.60, 396.9…
## $ lstat   <dbl> 4.98, 9.14, 4.03, 2.94, 5.33, 5.21, 12.43, 19.15, 29.93, 17.1…
## $ medv    <dbl> 24.0, 21.6, 34.7, 33.4, 36.2, 28.7, 22.9, 27.1, 16.5, 18.9, 1…
summary(Boston)
##       crim                zn             indus            chas        
##  Min.   : 0.00632   Min.   :  0.00   Min.   : 0.46   Min.   :0.00000  
##  1st Qu.: 0.08204   1st Qu.:  0.00   1st Qu.: 5.19   1st Qu.:0.00000  
##  Median : 0.25651   Median :  0.00   Median : 9.69   Median :0.00000  
##  Mean   : 3.61352   Mean   : 11.36   Mean   :11.14   Mean   :0.06917  
##  3rd Qu.: 3.67708   3rd Qu.: 12.50   3rd Qu.:18.10   3rd Qu.:0.00000  
##  Max.   :88.97620   Max.   :100.00   Max.   :27.74   Max.   :1.00000  
##       nox               rm             age              dis        
##  Min.   :0.3850   Min.   :3.561   Min.   :  2.90   Min.   : 1.130  
##  1st Qu.:0.4490   1st Qu.:5.886   1st Qu.: 45.02   1st Qu.: 2.100  
##  Median :0.5380   Median :6.208   Median : 77.50   Median : 3.207  
##  Mean   :0.5547   Mean   :6.285   Mean   : 68.57   Mean   : 3.795  
##  3rd Qu.:0.6240   3rd Qu.:6.623   3rd Qu.: 94.08   3rd Qu.: 5.188  
##  Max.   :0.8710   Max.   :8.780   Max.   :100.00   Max.   :12.127  
##       rad              tax           ptratio          black       
##  Min.   : 1.000   Min.   :187.0   Min.   :12.60   Min.   :  0.32  
##  1st Qu.: 4.000   1st Qu.:279.0   1st Qu.:17.40   1st Qu.:375.38  
##  Median : 5.000   Median :330.0   Median :19.05   Median :391.44  
##  Mean   : 9.549   Mean   :408.2   Mean   :18.46   Mean   :356.67  
##  3rd Qu.:24.000   3rd Qu.:666.0   3rd Qu.:20.20   3rd Qu.:396.23  
##  Max.   :24.000   Max.   :711.0   Max.   :22.00   Max.   :396.90  
##      lstat            medv      
##  Min.   : 1.73   Min.   : 5.00  
##  1st Qu.: 6.95   1st Qu.:17.02  
##  Median :11.36   Median :21.20  
##  Mean   :12.65   Mean   :22.53  
##  3rd Qu.:16.95   3rd Qu.:25.00  
##  Max.   :37.97   Max.   :50.00

The Boston data frame has 506 rows and 14 columns. All data is numeric. Variables are continuous expect binary variable chas, (“1 if tract bounds Charles River; 0 otherwise”). Response/target variable will be crim (“per capita crime rate by town”), which has range 0.06-89.0 and mean=3.6 & median=0.26. Difference between mean and median suggests strongly skewed distribution.

3. Graphical overview and summaries

Descriptives were already shown in step 2. Variable distributions have clearly distinct variances and locations. Some variables seem to be strongly skewed as well. Let’s have a visual.

# Bivariate scatter plots (excluding 4th variable chas)

par(mfrow = c(1,2), pin=c(1.75,1.75))
hist(Boston$crim,col="blue",xlab=NULL)
boxplot(Boston$crim,col="blue", main="Boxplot of Boston$crim")

We can see that crime distribution is strongly skewed. We are going to standardize and categorize variables later.

Pairwise distributions (the plot is enlarged with defining additional R-chunk parameters, which are not shown in output):

# Bivariate scatter plots (excluding 4th variable chas)
pairs(Boston[,c(1:3,5:dim(Boston)[2])], pch=19, cex=0.05, lower.panel=NULL)

Scatter plot matrix shows that bivariate distributions are not always multinormal (or almost never, to be honest). Some distributions do have large empty areas and/or outlier observations, like rad (index of accessibility to radial highways). Not good.

Some mutual bivariate associations seems to be stronger than others. Let’s calculate correlations.

# calculate the correlation matrix and round it
cor_matrix<-cor(Boston) %>% round(digits=2)

# print the correlation matrix
cor_matrix
##          crim    zn indus  chas   nox    rm   age   dis   rad   tax ptratio
## crim     1.00 -0.20  0.41 -0.06  0.42 -0.22  0.35 -0.38  0.63  0.58    0.29
## zn      -0.20  1.00 -0.53 -0.04 -0.52  0.31 -0.57  0.66 -0.31 -0.31   -0.39
## indus    0.41 -0.53  1.00  0.06  0.76 -0.39  0.64 -0.71  0.60  0.72    0.38
## chas    -0.06 -0.04  0.06  1.00  0.09  0.09  0.09 -0.10 -0.01 -0.04   -0.12
## nox      0.42 -0.52  0.76  0.09  1.00 -0.30  0.73 -0.77  0.61  0.67    0.19
## rm      -0.22  0.31 -0.39  0.09 -0.30  1.00 -0.24  0.21 -0.21 -0.29   -0.36
## age      0.35 -0.57  0.64  0.09  0.73 -0.24  1.00 -0.75  0.46  0.51    0.26
## dis     -0.38  0.66 -0.71 -0.10 -0.77  0.21 -0.75  1.00 -0.49 -0.53   -0.23
## rad      0.63 -0.31  0.60 -0.01  0.61 -0.21  0.46 -0.49  1.00  0.91    0.46
## tax      0.58 -0.31  0.72 -0.04  0.67 -0.29  0.51 -0.53  0.91  1.00    0.46
## ptratio  0.29 -0.39  0.38 -0.12  0.19 -0.36  0.26 -0.23  0.46  0.46    1.00
## black   -0.39  0.18 -0.36  0.05 -0.38  0.13 -0.27  0.29 -0.44 -0.44   -0.18
## lstat    0.46 -0.41  0.60 -0.05  0.59 -0.61  0.60 -0.50  0.49  0.54    0.37
## medv    -0.39  0.36 -0.48  0.18 -0.43  0.70 -0.38  0.25 -0.38 -0.47   -0.51
##         black lstat  medv
## crim    -0.39  0.46 -0.39
## zn       0.18 -0.41  0.36
## indus   -0.36  0.60 -0.48
## chas     0.05 -0.05  0.18
## nox     -0.38  0.59 -0.43
## rm       0.13 -0.61  0.70
## age     -0.27  0.60 -0.38
## dis      0.29 -0.50  0.25
## rad     -0.44  0.49 -0.38
## tax     -0.44  0.54 -0.47
## ptratio -0.18  0.37 -0.51
## black    1.00 -0.37  0.33
## lstat   -0.37  1.00 -0.74
## medv     0.33 -0.74  1.00

These are Pearson correlation coefficients, which are not always reliable in a case of non-normality. But let’s keep them anyway, like in Datacamp. However, it’s challenging to assimilate so many numbers. Better to have a visualization.

# installing and/or loading corrplot library for correlation coefficient visualization
if (!require("corrplot")) {
   install.packages("corrplot")
   library(corrplot)
}
## Loading required package: corrplot
## corrplot 0.84 loaded
# visualize the correlation matrix
corrplot.mixed(cor_matrix, tl.cex=0.75, number.cex=0.75, number.digits=2)

That’s fancy way to present correlation coefficient. We can directly see that there are some strong correlations indicated by big spheres, blue ones for negative and red ones for positive values. Highest value is +0.91, which is between rad and tax (index of accessibility to radial highways and full-value property-tax rate per $10,000).

4. Standardization and categorization

# center and standardize variables
# 'as.data.frame' is needed since it makes referring variables easier later
boston_scaled <- as.data.frame(scale(Boston))

# Bivariate scatter plots
pairs(boston_scaled, pch=19, cex=0.05, lower.panel=NULL)

#Means, second argument refers to columns
apply(boston_scaled, 2, FUN=mean)
##          crim            zn         indus          chas           nox 
## -7.202981e-18  2.282481e-17  1.595296e-17 -3.544441e-18 -2.150022e-16 
##            rm           age           dis           rad           tax 
## -1.056462e-16 -1.643357e-16  1.153079e-16  4.799652e-17  2.024415e-17 
##       ptratio         black         lstat          medv 
## -3.924246e-16 -1.151679e-16 -7.052778e-17 -1.374631e-16
#Variances, second argument refers to columns
apply(boston_scaled, 2, FUN=var)
##    crim      zn   indus    chas     nox      rm     age     dis     rad     tax 
##       1       1       1       1       1       1       1       1       1       1 
## ptratio   black   lstat    medv 
##       1       1       1       1

We can see that now all variables has been standardized, i.e. scaled to have mean=0 (or very close to it, at least) and variance=1. This can’t heal difficult distributions like rad or chas has, though. Correlations are the same, so they are not shown here again.

Note: For some reason scaling must be for R data.frame object, otherwise further operations won’t work. This means that command
    boston_scaled <- scale(Boston)
is not enought, but I need to put it this way:
    boston_scaled <- as.data.frame(scale(Boston)).
So strange… And the same applies for task #7.

Categorization

Creating categorized version of crime variable. Cut points are quantiles, so number of categories will be four.

# create a quantile vector of crim and print it
bins <- quantile(boston_scaled$crim)
bins
##           0%          25%          50%          75%         100% 
## -0.419366929 -0.410563278 -0.390280295  0.007389247  9.924109610
# create a categorical variable 'crime'
crime <- cut(boston_scaled$crim, breaks=bins, include.lowest=TRUE, labels=c("low","med_low","med_high","high"))

# look at the table of the new factor crime
table(crime)
## crime
##      low  med_low med_high     high 
##      127      126      126      127

Looks fine, distribution is as close to 25/25/25/25 percentages as possible.

# remove original crim from the dataset
boston_scaled <- dplyr::select(boston_scaled, -crim)

# add the new categorical value to scaled data
boston_scaled <- data.frame(boston_scaled, crime)

Now original crim has been replaced to categorical crime.

Dividing the dataset to train and test sets

# number of rows in the Boston dataset 
n <- nrow(boston_scaled)

# choose randomly 80% of the rows
ind <- sample(n,  size = n * 0.8)

# create train set
train <- boston_scaled[ind,]

# create test set 
test <- boston_scaled[-ind,]

dim(train)
## [1] 404  14
dim(test)
## [1] 102  14

Now datasets train and test has been created, with the first comprising 80% and latter 20% of original cases.

5. Linear discriminant analysis

Fitting LDA model on the train set with crime rate categorizations as target variable. Other variables are used as predictors.

# linear discriminant analysis
lda.fit <- lda(crime ~ ., data = train)

# print the lda.fit object
lda.fit
## Call:
## lda(crime ~ ., data = train)
## 
## Prior probabilities of groups:
##       low   med_low  med_high      high 
## 0.2450495 0.2500000 0.2425743 0.2623762 
## 
## Group means:
##                  zn      indus        chas        nox         rm        age
## low       0.9155784 -0.8724720 -0.07348562 -0.8641916  0.4296231 -0.8648879
## med_low  -0.1108978 -0.2897329  0.03952046 -0.5696773 -0.1631957 -0.3321327
## med_high -0.3804849  0.1280289  0.08924136  0.3494963  0.0388755  0.4095427
## high     -0.4872402  1.0170298 -0.04947434  1.0545704 -0.4538456  0.8088602
##                 dis        rad        tax     ptratio      black       lstat
## low       0.8279540 -0.7069352 -0.7423643 -0.46282839  0.3778766 -0.75865593
## med_low   0.3381695 -0.5429521 -0.4674443 -0.03708457  0.3514022 -0.10433231
## med_high -0.3316753 -0.4228725 -0.3307720 -0.25236197  0.1489785 -0.01024572
## high     -0.8547089  1.6390172  1.5146914  0.78181164 -0.7629361  0.82667708
##                medv
## low       0.5370072
## med_low  -0.0123946
## med_high  0.1290169
## high     -0.6336874
## 
## Coefficients of linear discriminants:
##                  LD1         LD2         LD3
## zn       0.095289166  0.77020974 -0.96026519
## indus    0.017428409 -0.19918318  0.24396499
## chas    -0.008251035 -0.01373677  0.12016330
## nox      0.419779620 -0.76069570 -1.36610297
## rm      -0.009432696 -0.08246193 -0.17085193
## age      0.219970998 -0.46587895 -0.19795373
## dis     -0.058372705 -0.35540296  0.14414717
## rad      3.349715907  0.78986897 -0.06082624
## tax      0.030766675  0.14169669  0.60286798
## ptratio  0.113005513  0.07162020 -0.26477437
## black   -0.132015277  0.03192794  0.18379144
## lstat    0.146366063 -0.13765569  0.52581600
## medv     0.073741014 -0.28585033 -0.09675911
## 
## Proportion of trace:
##    LD1    LD2    LD3 
## 0.9566 0.0313 0.0121
classes <- as.numeric(train$crime)

# LDA (bi)plot
plot(lda.fit, col=classes)

Interestingly plot() function adds group names as data value symbols independently on pch parameters. For example, following commands are yielding identical plots:
- plot(lda.fit, col=classes, pch=19)
- plot(lda.fit, col=classes, pch=classes)
I have no clue why. Showing category value as symbols looks very awkward.

Anyway, plot visualizes how target variable classes are separated by the linear combinations of predictor variables. Looks like crime quantile groups are mostly very nicely separated. High crime group seems to be most consistent while med_high is most scattered around.

Let’s enhance plot by adding arrows into it, just like in Datacamp exercise.

# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
  heads <- coef(x)
  arrows(x0 = 0, y0 = 0, 
         x1 = myscale * heads[,choices[1]], 
         y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
  text(myscale * heads[,choices], labels = row.names(heads), 
       cex = tex, col=color, pos=3)
}

plot(lda.fit, col=classes, pch=classes, dimen=2)
lda.arrows(lda.fit, myscale = 2)

Looks like rad (“index of accessibility to radial highways”) is an important factor here. Watch out for living close to radial highways!

6. Predicting crime with LDA model

# save the correct classes from test data
correct_classes <- test$crime

# remove the crime variable from test data
test <- dplyr::select(test, -crime)

# predict classes with test data
lda.pred <- predict(lda.fit, newdata = test)

# cross tabulate the results
table(correct = correct_classes, predicted = lda.pred$class)
##           predicted
## correct    low med_low med_high high
##   low       18       9        1    0
##   med_low    6      11        8    0
##   med_high   0       4       22    2
##   high       0       0        0   21

Cross-tabulation shows that in 13+17+15+20 cases (which is 65) prediction is fully correct, leaving out 37 more or less incorrect cases. Proportion of correct predictions is hence 0.6372549. We can see that if prediction has gone wrong, it has still has been mostly placed into category close to the real one. And, we can see that all 20 high crimes cases are correctly predicted. Not bad at all, I guess.

7. Distances, K-means clustering

Euclidian distances

# reloading the Boston dataset 
data('Boston')

# standardizing the dataset
# again, as.data.frame is needed, otherwise further operations won't work.
boston_scaled <- as.data.frame(scale(Boston))

# euclidean and manhattan distance matrix
dist_eu <- dist(boston_scaled)
dist_man <- dist(boston_scaled, method="manhattan")

# look at the summaries of the distances
summary(dist_man)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##  0.2662  8.4832 12.6090 13.5488 17.7568 48.8618
summary(dist_man)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##  0.2662  8.4832 12.6090 13.5488 17.7568 48.8618
par(mfrow = c(2,2), pin=c(1.75,1.75))
#Euclidean
hist(dist_eu,col="blue",xlab=NULL)
boxplot(dist_eu,col="blue", main="Boxplot of dist_eu")
#Manhattan
hist(dist_man,col="blue",xlab=NULL)
boxplot(dist_man,col="blue", main="Boxplot of dist_man")

Mean value for Euclidean distance is around 5, range is 0.1 - 14.4. Distribution is slightly skewed to the right.

Mean value for Manhattan distance is around 14, range is 0.3 - 48.9. Distribution is a bit more skewed to the right.

K-means clustering

Let’s conduct k-means clustering with three clusters, which, I think, might be good low-but-not-super-low number for centers/clusters.

# k-means clustering with three clusters (=semirandomly selected number)
km <-kmeans(boston_scaled, centers=3)

# plot the Boston dataset with clusters
pairs(boston_scaled, col=km$cluster, pch=19, cex=0.05, lower.panel=NULL)

Three cluster solutions might be plausible. Groups are formed nicely. But let’s see which is optimal number based on calculations.

# Setting a seed for random generator
set.seed(322654435)

# determine the number of clusters
k_max <- 10

# calculate the total within sum of squares
twcss <- sapply(1:k_max, function(k){kmeans(boston_scaled, k)$tot.withinss})

# visualize the results
library(ggplot2)
ggplot() + 
  geom_line(aes(x=1:k_max, y=twcss)) +
  scale_x_continuous(breaks=c(1:10)) +
  xlab("Number of cluster")

If the principle is that the optimal number of clusters is when the value of total WCSS changes radically, then two clusters would be good choice now. After two clusters total within sum of squares decreased slower. So let’s conduct k-means clustering again, now with two centers, and visulize results.

# k-means clustering
km2 <-kmeans(boston_scaled, centers=2)

# plot the Boston dataset with clusters
pairs(boston_scaled, col=km2$cluster, pch=19, cex=0.05, lower.panel=NULL)

This looks plausible as well. Dot colors indicating cluster seem to mostly create distinct groups in these subplots.

Bonus task

Reloading and scaling Boston data set. AFAIK, this is not needed to done again, but let’s follow instructions. K-means clustering with three center. Fitting LDA with cluster as a target variable, keeping all Boston variables as predictors. Adding arrows with previously created custom function, replacing default red color with blue for better definition. Aspect ratio, text size and arrow length scaling changed from default values as well.

# reloading the Boston dataset 
data('Boston')
boston_scaled <- as.data.frame(scale(Boston))

# k-means clustering with three centers
km3 <- kmeans(boston_scaled, centers=3)
km3
## K-means clustering with 3 clusters of sizes 213, 129, 164
## 
## Cluster means:
##         crim         zn      indus         chas        nox         rm
## 1 -0.3792002 -0.3640439 -0.2654730  0.004931512 -0.3761109 -0.2611171
## 2 -0.3978261  1.2205329 -0.9803713 -0.028167813 -0.8262430  1.0139959
## 3  0.8054220 -0.4872402  1.1159370  0.015751437  1.1383961 -0.4584605
##           age        dis        rad        tax     ptratio      black
## 1 -0.06380464  0.1123306 -0.5877259 -0.5858342  0.04591154  0.2799627
## 2 -0.90963302  0.9031496 -0.5839140 -0.6780000 -0.82270661  0.3543735
## 3  0.79837224 -0.8562971  1.2226252  1.2941749  0.58749997 -0.6423552
##         lstat       medv
## 1 -0.09228022 -0.1222510
## 2 -0.94791366  1.0675487
## 3  0.86546676 -0.6809409
## 
## Clustering vector:
##   1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16  17  18  19  20 
##   2   1   2   2   2   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1 
##  21  22  23  24  25  26  27  28  29  30  31  32  33  34  35  36  37  38  39  40 
##   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   2 
##  41  42  43  44  45  46  47  48  49  50  51  52  53  54  55  56  57  58  59  60 
##   2   2   1   1   1   1   1   1   1   1   1   1   2   2   2   2   2   2   2   1 
##  61  62  63  64  65  66  67  68  69  70  71  72  73  74  75  76  77  78  79  80 
##   1   1   1   2   2   2   2   1   1   1   1   1   1   1   1   1   1   1   1   1 
##  81  82  83  84  85  86  87  88  89  90  91  92  93  94  95  96  97  98  99 100 
##   2   1   2   1   1   1   1   1   1   1   1   1   1   1   1   1   1   2   2   2 
## 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 
##   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1 
## 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 
##   1   1   1   1   1   1   1   3   3   3   1   1   1   1   3   3   3   3   3   3 
## 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 
##   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   1   1   3 
## 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 
##   1   2   2   2   1   1   2   1   1   1   1   1   1   1   1   1   1   1   1   2 
## 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 
##   2   1   2   1   1   1   2   2   2   2   2   2   2   2   2   2   2   2   2   2 
## 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 
##   2   2   2   2   2   1   1   1   1   1   1   1   1   1   1   1   1   1   1   1 
## 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 
##   1   1   1   1   2   2   2   1   2   2   1   2   2   2   1   1   1   2   2   2 
## 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 
##   2   1   2   2   1   1   2   1   2   2   2   2   2   2   2   2   2   2   2   2 
## 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 
##   2   2   2   2   2   1   2   2   2   1   1   2   1   2   2   2   2   2   2   2 
## 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 
##   2   2   2   2   2   2   2   2   2   2   2   2   2   1   1   2   1   1   2   2 
## 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 
##   2   2   2   2   2   2   2   2   1   1   1   1   1   1   1   1   1   1   1   1 
## 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 
##   1   1   1   1   1   1   1   1   1   1   1   1   2   1   1   1   1   1   1   1 
## 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 
##   1   2   1   2   2   1   1   2   2   2   2   2   2   2   2   2   3   3   3   3 
## 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 
##   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3 
## 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400 
##   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3 
## 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 
##   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3 
## 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 
##   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3 
## 441 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 459 460 
##   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3 
## 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 478 479 480 
##   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3   3 
## 481 482 483 484 485 486 487 488 489 490 491 492 493 494 495 496 497 498 499 500 
##   3   3   3   3   3   3   3   3   3   3   3   3   3   1   1   1   1   1   1   1 
## 501 502 503 504 505 506 
##   1   1   1   1   1   1 
## 
## Within cluster sum of squares by cluster:
## [1] 1083.594 1064.601 1727.469
##  (between_SS / total_SS =  45.2 %)
## 
## Available components:
## 
## [1] "cluster"      "centers"      "totss"        "withinss"     "tot.withinss"
## [6] "betweenss"    "size"         "iter"         "ifault"
# LDA with cluster as a target
lda2.fit <- lda(km3$cluster ~ ., data=boston_scaled)

# print the lda.fit object
lda2.fit
## Call:
## lda(km3$cluster ~ ., data = boston_scaled)
## 
## Prior probabilities of groups:
##         1         2         3 
## 0.4209486 0.2549407 0.3241107 
## 
## Group means:
##         crim         zn      indus         chas        nox         rm
## 1 -0.3792002 -0.3640439 -0.2654730  0.004931512 -0.3761109 -0.2611171
## 2 -0.3978261  1.2205329 -0.9803713 -0.028167813 -0.8262430  1.0139959
## 3  0.8054220 -0.4872402  1.1159370  0.015751437  1.1383961 -0.4584605
##           age        dis        rad        tax     ptratio      black
## 1 -0.06380464  0.1123306 -0.5877259 -0.5858342  0.04591154  0.2799627
## 2 -0.90963302  0.9031496 -0.5839140 -0.6780000 -0.82270661  0.3543735
## 3  0.79837224 -0.8562971  1.2226252  1.2941749  0.58749997 -0.6423552
##         lstat       medv
## 1 -0.09228022 -0.1222510
## 2 -0.94791366  1.0675487
## 3  0.86546676 -0.6809409
## 
## Coefficients of linear discriminants:
##                  LD1          LD2
## crim    -0.032231372 -0.180555401
## zn       0.009313889 -1.068645925
## indus    0.624151354 -0.006677424
## chas     0.039930549  0.124911531
## nox      1.097435166 -0.777946957
## rm      -0.193309184 -0.586124321
## age     -0.164347859  0.406286594
## dis      0.052315163 -0.286780919
## rad      0.701339082 -0.182828969
## tax      1.041861064 -0.515465919
## ptratio  0.245758857  0.034929672
## black   -0.020655070  0.015978407
## lstat    0.187227781 -0.395217119
## medv    -0.083275066 -0.768211713
## 
## Proportion of trace:
##    LD1    LD2 
## 0.8504 0.1496
# target classes as numeric
classes <- as.numeric(km3$cluster)

# plot the lda results
plot(lda2.fit, col=classes, pch=classes)

# Using previously defined function for lda biplot arrows
lda.arrows(lda2.fit, color="blue", tex=0.8, myscale=5)

Looks like variables nox (nitrogen oxides concentration, parts per 10 million), tax (full-value property-tax rate per $10,000) and zn (proportion of residential land zoned for lots over 25,000 sq.ft) are the most influental separators for the clusters among all variables, based on arrow lengths.

Super-Bonus task

Running the given code for the scaled train data. The code creates a matrix product, which is a projection of the data points. Installing and loading plotly package and creating 3D plots.

Adjusting the code. Defining symbol color as train data set crime class. Drawing another 3D plot where the color is defined by the clusters of the k-means. Plots are only shown in RStudio viewer, so no output in course diary.

# creating train set (again)
train <- boston_scaled[ind,]

#K-means clustering with three centers:
km4 <- kmeans(train, centers=3)

#restoring original crime classification into train data
crime <- cut(train$crim, breaks=bins, include.lowest=TRUE, labels=c("low","med_low","med_high","high"))

#model_predictors <- dplyr::select(train, everything())
model_predictors <- dplyr::select(train, c(-crim))
#dim(model_predictors)

# check the dimensions
dim(model_predictors)
## [1] 404  13
dim(lda.fit$scaling)
## [1] 13  3
# matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)

# installing and/or loading plotly package for 3D plot
if (!require("plotly")) {
   install.packages("plotly")
   library(plotly)
}
## Loading required package: plotly
## 
## Attaching package: 'plotly'
## The following object is masked from 'package:MASS':
## 
##     select
## The following object is masked from 'package:ggplot2':
## 
##     last_plot
## The following object is masked from 'package:stats':
## 
##     filter
## The following object is masked from 'package:graphics':
## 
##     layout
#Create a 3D plot (Cool!) of the columns of the matrix product by typing the code below.
#plot_ly(x=matrix_product$LD1, y=matrix_product$LD2, z=matrix_product$LD3, type='scatter3d', mode='markers')

#Graph 1
plot_ly(x=matrix_product$LD1, y=matrix_product$LD2, z=matrix_product$LD3, color=crime, type='scatter3d', mode='markers') 
## Warning: `arrange_()` is deprecated as of dplyr 0.7.0.
## Please use `arrange()` instead.
## See vignette('programming') for more help
## This warning is displayed once every 8 hours.
## Call `lifecycle::last_warnings()` to see where this warning was generated.
#Graph 2
plot_ly(x=matrix_product$LD1, y=matrix_product$LD2, z=matrix_product$LD3, color=as.factor(km4$cluster), type='scatter3d', mode='markers')

How do plots differ? Well, data points are identical in x-y-z space. Only number of categories/colors (4 vs. 3) differ, so do group distribution. After all, both plots are telling pretty much the same story: there is one clearly distinct blob (areas with highest crime rates) while rest of dots are somewhat grouped as well. Maybe there is bit more overlapping (variation) in crime category plot.